Definition: Let $(X, \mathcal{M}, \mu)$ a measure space. Some property P (in this case, the continuity) is said to be satisfied almost everywhere in X if there exists a set $N \in \mathcal{M}$ such that $\mu(N) = 0$ and for all $x \in X\setminus N$, the property P holds.
Problem: Let $f,g:\mathbb{R}\rightarrow\mathbb{R}$ functions and g continuous in $\mathbb{R}$ with $g\neq0$. The function defined
$$f(x) = \begin{cases}
g(x) & x \in \mathbb{R}\setminus\mathbb{Q} \\
0 & x \in \mathbb{Q}
\end{cases}
$$
prove that $f$ is $\mathcal{L}_{a.e}$ continuous (that is, f is continuous almost-everywhere using the Lebesgue measure).
Is there an easy way to prove this?
Best Answer
If $E=\mathbb Q$ then $E$ has measure $0$ and the restriction of $f$ to $\mathbb R \setminus E$ is $g$ which is continuous.
You cannot prove the existence of a set $E$ of measure $0$ such that $f$ is continuous at each point of $\mathbb R \setminus E$. For example, if $g(x)=1+|x|$ then the corresponding $f$ is not continuous at any point!. If your null set exists then we get the conclusion that the real line has measure $0$!.